JEE Binomial Theorem Previous Year Questions With Solutions

The binomial theorem describes the coefficients of each term when we expand (a + b)n. When n is equal to a prime number p, p will divide all those coefficients, except the end one’s. Binomial theorem is one of the important concepts in JEE Mathematics. This theorem also plays a prominent role to proof other results or theorems. About 1-2 question/s asked from this topic in JEE Examination. The subject experts at BYJU’S bring chapter-wise previous year solved questions of Binomial Theorem including important concepts and formulae to help JEE aspirants. The study material given in this article can be used for a quick revision before the examination. Students can freely download the solutions and start practising offline.

Download Binomial Theorem Previous Year Solved Questions PDF

 Important concepts to Remember:

1. Definition: Let a and b be numbers and n be a natural number, then

(a+b)n=i=0n(ni)anibi(a + b)n = \sum_{i=0}^n\binom{n}{i} a^{n-i}b^i

2. Formula to find the coefficient from Pascal’s Triangle:

(ni)=n!k!(nk)!\binom{n}{i} = \frac{n!}{k!(n-k)!}

3. The binomial theorem fails arithmetically when it expands a finite power of a binomial in an infinite divergent series.

JEE Main Past Year Questions With Solutions on Binomial Theorem

Question 1: If the sum of the coefficients of all even powers of x in the product (1 + x + x2 + … + x2n) (1 – x + x2 – x3 + … + x2n) is 61, then find the value of n.

Solution:

Let (1 + x + x2 + … + x2n) (1 – x + x2 – x3 + … + x2n) = a0 + a1x + a2x2 + a3x3+ ……

Put x = 1 => 2n + 1 = a0 + a1 + a2 + a3 + …… …(i)

Put x = -1 => 2n + 1 = a0 – a1 + a2 – a3 + …… …..(ii)

Add (i) and (ii), we get

2n + 1 = 61

Or n = 30

Question 2: If α and β be the coefficients of x4 and x2 respectively in the expansion of (x + √(x2 − 1))6 + (x − √(x2 − 1)6, then:

(a) α + β = −30

(b) α− β = −132

(c) α− β = 60

(d) α+ β = 60

Answer: (b)

Solution: (x + √(x2 − 1))6 + (x − √(x2 − 1)6

= 2[6C0 x6 + 6C2x4(x2 – 1) + 6C4x2(x2 – 1)2 + 6C6(x2 – 1)3]

= 2[32x6 – 48x4 + 18x2 – 1]

=> α = -96 and β = 36

=> α – β = -132

Question 3: Find the coefficient of x4 in the expansion of (1 + x + x2)10 .

Solution:

General term of the given expression is,

10!p!q!r!xq+2r\frac{10!}{p!q!r!} x^{q + 2r}

Here, q + 2r = 4

For p = 6, q = 4, r = 0, coefficient = 10!/[6! x 4!] = 210

For p = 7, q = 2, r = 1, coefficient = 10! /[7! x 2! x 1!] = 360

For p = 8, q = 0, r = 2, coefficient = 10!/[ 8! x 2!] = 45

Therefore, sum = 615

Question 4: If x is positive, the first negative term in the expansion of

(1 + x)27/5 is

(a) 5th term (b) 6th term (c) 7th term (d) 8th term

Answer: (d)

Solution:

We know, Tr+1=n(n1)(nr+1)r!xrT_{r+1} = \frac{n(n-1)…(n-r+1)}{r!} x^r

Therefore, n – r + 1 < 0

=> 27/5 + 1 < r

or r > 6

Here r = 7

So, Tr+1 < 0

Hence Tr+1 = T7+1 = T8 = 8th term is negative.

Question 5: The positive integer just greater than (1 + 0.0001)10000 is

(a) 4 (b) 5 (c) 2 (d) 3

Answer: (d)

Solution:

(1 + 0.0001)10000 which is similar to the form (1 + 1/n)n where n = 10000.

Using binomial expansion, we have

(1 + 0.0001)10000 = (1 + 1/n)n

= 1 + n x 1/n + n(n-1)/2! x 1/n2 + [n(n-1)(n-2)]/3! X 1/n3 + ……..

= 1 + 1 + 1/2!(1 – 1/n) + 1/3!(1 – 1/n) + (1 – 2/n) + ……

< 1 + 1/1! + 1/2! + 1/3! + …..+1/(9999)!

= 1 + 1/1! + 1/2! + 1/3! + …….∞

= e < 3

Question 6: The coefficients of xp and xq in the expression of (1 + x)p+q are

(a) Equal

(b) Equal with opposite signs

(c) Reciprocals of each other

(d) None of these

Answer: (a)

Solution:

The coefficients of xp and xq in the expression of (1 + x)p+q

We know, yp+1 = p+qCp xp and yq+1 = p+qCq xq

Coefficient of xp = p+qCp = [p+q]!/p!q!

and

Coefficient of xq = p+qCq = [p+q]!/p!q!

Therefore, coefficients of xp and xq in the expression of (1 + x)p+q are equal.

Question 7: If an=7+7+7+a_n = \sqrt{7+\sqrt{7+\sqrt{7+…}}} having n radical signs then by methods of mathematical induction which is true

(a) an < 7 for all n ≥ 1

(b) an > 7 for all n ≥ 1

(c) an > 3 for all n ≥ 1

(d) an < 4 for all n ≥ 1

Answer: (c)

Solution:

an=7+7+7+a_n = \sqrt{7+\sqrt{7+\sqrt{7+…}}}

=>an = √(7+an)

=> an2 – an – 7 = 0

Solving above quadratic equation, we get

an = [1±√29]/2

But an > 0, therefore, an = [1+√29]/2 > 3

=> an > 3

Question 8: If the sum of the coefficients in the expansion of (a + b)n is 4096, then the greatest coefficient in the expansion is

(a) 1594 (b) 924 (c) 792 (d) 2924

Answer: (b)

Solution:

Binomial expansion: (x + y)n = C0 xn + C1 xn-1 y + C2 xn-2y2 + ….+ Cnyn

Putting x = y = 1, we get

2n = C0 + C1 + C2 + ….+ Cn = 4096 = 212

=> n = 12

Since n is even here, so the coefficient of greatest term is

nCn/2 = 12C12/2 = 12C6

= 12/6 x 11/5 x 10/4 x 9/3 x8/2 x 7/1

= 924

Question 9: The sum of the co-efficient of all odd degree terms in the expansion of

(x + √(x3-1)5 + (x – √(x3-1)5 , (x > 1) is

(a) 0 (b) 2 (c) 1 (d) -1

Answer:

Solution:

Let y = √(x3-1)

So, given expression reduced as (x + y)5 + (x – y)5

Using binomial expansion, we have

(x + y)5 + (x – y)5 = x5 + 5C1 x4y + 5C2 x3y2 + 5C3 x2y3 + 5C4 xy4 + 5C5 y5 + x55C1 x4y + 5C2 x3y25C3 x2y3 + 5C4 xy45C5 y5

= 2(x5 + 10 x3y2 + 5xy4)

= 2(x5 + 10 x3(x3 – 1 ) + 5x (x3 – 1)2)

= 2{10x6 + 5x7 + x5 – 10x3 – 10x4 + 5x}

Sum of coefficient of odd powers of x: 2{5 + 1 – 10 + 5} = 2

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